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### Section 6-4 : Surface Integrals of Vector Fields

1. Evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$$ where $$\vec F = 3x\,\vec i + 2z\,\vec j + \left( {1 - {y^2}} \right)\vec k$$ and $$S$$ is the portion of $$z = 2 - 3y + {x^2}$$ that lies over the triangle in the xy-plane with vertices $$\left( {0,0} \right)$$, $$\left( {2,0} \right)$$ and $$\left( {2, - 4} \right)$$ oriented in the negative $$z$$-axis direction. Solution
2. Evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$$ where $$\vec F = - x\,\vec i + 2y\,\vec j - z\,\vec k$$ and $$S$$ is the portion of $$y = 3{x^2} + 3{z^2}$$ that lies behind $$y = 6$$ oriented in the positive $$y$$-axis direction. Solution
3. Evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$$ where $$\vec F = {x^2}\,\vec i + 2z\,\vec j - 3y\,\vec k$$ and $$S$$ is the portion of $${y^2} + {z^2} = 4$$ between $$x = 0$$ and $$x = 3 - z$$ oriented outwards (i.e. away from the $$x$$-axis). Solution
4. Evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$$ where $$\vec F = \,\vec i + z\,\vec j + 6x\,\vec k$$ and $$S$$ is the portion of the sphere of radius 3 with $$x \le 0$$, $$y \ge 0$$ and $$z \ge 0$$ oriented inward (i.e. towards the origin). Solution
5. Evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$$ where $$\vec F = y\,\vec i + 2x\,\vec j + \left( {z - 8} \right)\,\vec k$$ and $$S$$ is the surface of the solid bounded by $$4x + 2y + z = 8$$, $$z = 0$$, $$y = 0$$ and $$x = 0$$ with the positive orientation. Note that all four surfaces of this solid are included in $$S$$. Solution
6. Evaluate $$\displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}$$ where $$\vec F = yz\,\vec i + x\,\vec j + 3{y^2}\,\vec k$$ and $$S$$ is the surface of the solid bounded by $${x^2} + {y^2} = 4$$, $$z = x - 3$$, and $$z = x + 2$$ with the negative orientation. Note that all three surfaces of this solid are included in $$S$$. Solution